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A water heater is fueled by the combustion of propane as described by the following reaction:
C3H8(g) + 5 O2(g) --> 3 CO2(g) + 4 H2O(l)

a) Determine how much heat is released if 11.2 moles of propane are burned at 298 K and at a constant pressure of 1.00 bar.

b) Assume that the water heater contains 146 L of water at an initial temperature of 25.0 degreeC. What will the water temperature rise to if all of the heat produced under the previous conditions is used to heat the water?

ANS:
a) -2.49x10^7 J
b) 65.8 degreeC

Please explain steps. Thank you! Much help is appreciated.

C3H8 + 5O2 ==> 3CO2 + 4H2O(l)

dHrxn = (n*dHf products) - (n*dHf reactants) and that x 11.2 for total dH.

b.
dH from above(+ sign here) = mass H2O x specific heat H2O x (Tfinal-Tinitial). Solve for Tf. I worked the problem and obtained those answers. By the way, thanks for including them in the post. I wish others would do that; it makes sure I don't drop a sign or forget a conversion.

Oh no problem :) and thank you so much for answering!

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